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You draw a card at random. Without replacing the first card, you draw a second card at random. What is the probability of drawing a $5$ and then a $6$ ? Write your answer as a fraction or whole number.

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Probability of independent and dependent events

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Q. You draw a card at random. Without replacing the first card, you draw a second card at random. What is the probability of drawing a

5

and then a

6

? Write your answer as a fraction or whole number.

Calculate Probability of Drawing $5$ : Determine the probability of drawing a $5$ from a standard deck of $52$ cards. $\newline$ There are $4$ fives in a deck of cards (one for each suit). The probability of drawing a $5$ is the number of fives divided by the total number of cards. $\newline$ $P(\text{Drawing a 5}) = \frac{\text{Number of fives}}{\text{Total number of cards}} = \frac{4}{52} = \frac{1}{13}$
Calculate Probability of Drawing $6$ : Determine the probability of drawing a $6$ after having drawn a $5$ , without replacement. $\newline$ After drawing a $5$ , there are now $51$ cards left in the deck. There are still $4$ sixes in the deck since we have not drawn a $6$ yet. The probability of drawing a $6$ from the remaining cards is the number of sixes divided by the remaining number of cards. $\newline$ $P(\text{Drawing a } 6 \text{ after a } 5) = \frac{\text{Number of sixes}}{\text{Remaining number of cards}} = \frac{4}{51}$
Calculate Combined Probability: Calculate the combined probability of both events happening in sequence (drawing a $5$ and then a $6$ without replacement). $\newline$ The combined probability is the product of the probabilities of each individual event. $\newline$ $P(\text{Drawing a 5 and then a 6}) = P(\text{Drawing a 5}) \times P(\text{Drawing a 6 after a 5}) = \frac{1}{13} \times \frac{4}{51}$
Simplify Final Probability: Simplify the expression to find the final probability. $\newline$ $P(\text{Drawing a } 5 \text{ and then a } 6) = \frac{1}{13} \times \frac{4}{51} = \frac{4}{13 \times 51} = \frac{4}{663}$

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